STRUCTURE OF IRIDIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( September 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I am with the student of Einstein Dr Th. Kalogeropoulos , who criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. There are two stable isotopes of iridium (Ir), the Ir-191 with S = +3/2 and the Ir-193 with S = =3/2. There are also 34 radioisotopes, the most stable radioisotope being Ir-192 with a half-life of 73.83 days, and many nuclear isomers, the most stable of which is Ir-192m with a half-life of 241 years, all other isomers have half-lives under a year, most under a day. Since the iridium is heavier than the osmium we conclude that it has 8 horizontal planes. Similarly all heavier nuclei than osmium have 8 horizontal planes able to give more blank positions than those of Osmium. For example the Pb-208 with 8 horizontal planes has 44 blank positions able to receive 44 extra neutrons with two bonds per neutron. You can see in Fig-7d the 8 horizontal planes of my paper “Nuclear structure is governed by the fundamental laws of electromagnetism”. ' '''The core of iridium, the Ir-154, with 77 protons and 77 neutrons (odd number) breaks the high symmetry giving only two stable isotopes. In general since the additional p77n77 is a vertical system with S =0, the structure of Ir-154 has S =0 with 8 horizontal planes of opposite spins . Moreover two horizontal squares like the -HSQ and +HSQ exist under and over the structure of the 8 horizontal planes. They give also S = 0 because the two deuterons of the down horizontal square (-HSQ) have S = -2 and the two deuterons of the up horizontal square ( +HSQ ) have S = +2. Of course several protons of such a core form blank positions able to receive 39 extra neutrons with two bonds per neutron for overcoming the pp and nn repulsions in the stable structure of the Ir-193. Moreover for constructing a stable structure of symmetrical arrangements in several nuclides as in the following group including the stable Ir-191 and Ir-193 the Ir-154 (core) changes the spin from S = 0 to S = +1 . In this case one deuteron of -HSQ changes the spin from S = -1 to S=0 giving S =+1 in order to form symmetrical arrangements with the additional p77n77. In other words for symmetrical arrangements the Ir-154 as a core has S = +1. Under this condition the stable Ir-193 with S = +3/2 of 39 extra neutrons has 20 extra neutrons of positive spins and 19 extra neutrons of negative spins. That is ' 'S = +1 + 20(+1/2) + 19(-1/2) = +3/2 ' 'On the other hand in the heavier unstable nuclides the more extra neutrons than those of the stable Ir-193 (in the absence of blank positions) make single bonds leading to the beta minus decay. ' ' ' 'STRUCTURE OF Ir-165, Ir-167, Ir-169, Ir-171, Ir-173, Ir-191, Ir-193, Ir-196, Ir-197, AND Ir-199 ' The structure of the above unstable nuclides including the stable structures of Ir-191 and Ir-193 is based also on the same structure of Ir-154 (core) having S = +1 . For example the unstable Ir-173 with S =+3/2 of 19 extra neutrons has 10 extra neutrons of positive spins and 9 extra neutrons of negative spins . That is S = +1 + 10(+1/2) + 9(-1/2) = +3/2 These extra neutrons fill the blank positions and make two bonds per neutron but their small number cannot give sufficient binding energies to pn bonds for overcoming the pp and nn repulsions. However in the stable structures of the Ir-191 and Ir-193 with S = +3/2 the greater number of extra neutrons gives enough binding energies to pn bonds for overcoming the repulsions. Whereas in the unstable Ir-195 with S = +3/2 the two more extra neutrons than those of the stable Ir-193 (in the absence of blank positions) make single bonds leading to the beta minus decay. In the same way the more extra neutrons than those of Ir-193 in the unstable nuclides like the Ir-196, Ir-197 and Ir-199 make single bonds leading to the beta minus decay. ' ' 'STRUCTURE OF Ir-172, Ir-174, Ir-180, Ir-182, Ir-186 AND Ir-192 ' After a careful analysis I found that the structure of this group is based on another structure of Ir-154 (core) having S = +3. In this case one deuteron of -HSQ changes the spin from S = -1 to S =+1 giving S = +2. Particularly it goes to +HSQ for making horizontal bonds with a deuteron of the up square. Also the additional p77n77 changes the spin from S = 0 to =+1. Particularly the vertical system of p77n77 with S =0 becomes a deuteron with S = +1 for making horizontal bonds with a deuteron of the +HSQ. Under this condition the unstable Ir-172 with S = +3 has 18 extra neutrons of opposite spins. Whereas the unstable Ir-192 with S=+4 of 38 extra neutrons has 20 extra neutrons of positive spins and 18 extra neutrons of negative spins. That is S = +3 + 20(+1/2) + 18(-1/2) = +4 '''STRUCTURE OF Ir-164, Ir-166, Ir-175, Ir-177, Ir-179, Ir-181, Ir-183, Ir-184, Ir-185, Ir-188, Ir-190, AND Ir-194 After a careful analysis I found that the structures of the above unstable nuclides are based on another structure of Ir-154 (core) having S = -2. In this case one deuteron of +HSQ changes the spin from S = +1 to S =-1 giving S = -2. Particularly it goes to -HSQ for making horizontal bonds with a deuteron of the down square. Under this condition the unstable Ir-164 with S = -2 has 10 extra neutrons of opposite spins. Whereas the unstable Ir-194 with S= -1 of 40 extra neutrons has 21 extra neutrons of positive spins and 19 extra neutrons of negative spins. That is S = -2 + 21(+1/2) + 19(-1/2) = -1 Category:Fundamental physics concepts